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The product rule is an essential tool in differential calculus when dealing with the derivative of a product of two functions. It's like a clever formula that saves us time and effort. Imagine we have two functions, say, \(u(x)\) and \(v(x)\), and we want to differentiate their product \(f(x) = u(x) \cdot v(x)\). The product rule tells us to take the derivative of the first function \(u(x)\) while keeping the second function \(v(x)\) as it is, then add the derivative of the second function times the first function kept as it is. In formula terms:

• \((uv)' = u'v + uv'\)

This rule helps us find the rate of change of the entire product, considering how each part changes on its own. In the exercise, we apply the product rule to find the derivative \(f'(x)\) of \(f(x) = x^2(3x-1)^{\frac{1}{3}}\). We identified \(u(x) = x^2\) and \(v(x) = (3x-1)^{\frac{1}{3}}\), finding both their derivatives first. The product rule then combines these to give the overall derivative of the original function.

The chain rule is another powerful technique in calculus used to differentiate composite functions. Sometimes, functions are nested within one another, like layers of an onion, and the chain rule helps us peel these layers to find the derivative. If we have a composite function \(y = g(f(x))\), where \(f(x)\) is the inner function and \(g(u)\) is the outer function, the chain rule states:

• \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)

This means we find how the outer function changes as its input changes, and multiply it by how the inner function changes.

In our exercise, we used the chain rule to differentiate \(v(x) = (3x-1)^{\frac{1}{3}}\). Here, the outer function is \(u^{1/3}\) and the inner function is \(3x-1\). By applying the chain rule, we found the derivative \(v'(x)\), which is part of applying the product rule later.

Derivatives are a fundamental concept in calculus that describe the rate at which a function is changing. In simple terms, if you think of a function as a journey, the derivative tells us the speed of this journey at any point. Imagine you have a car moving along a road: the derivative reveals how fast the car is going (its velocity) at each point along the road.

Using derivatives, we can solve problems across many fields such as physics, engineering, and economics. By determining derivatives, we can predict future behavior, optimize processes, and even understand how complex systems react to changes.

• The derivative of a function \(f(x)\) is often denoted as \(f'(x)\) or \(\frac{df}{dx}\).
• The derivative measures the function's instantaneous rate of change—that is, how it changes momentarily compared to a small change in \(x\).

In the exercise provided, we found the derivative of the function \(f(x) = x^2(3x-1)^{\frac{1}{3}}\) and evaluated it at \(x=3\), demonstrating the practical use of these concepts.